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10 tháng 3 2020

a) \(\frac{1}{10}-\frac{1}{40}-\frac{1}{88}-\frac{1}{154}-\frac{1}{238}-\frac{1}{340}\)

\(=\frac{1}{10}-\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)

\(=\frac{1}{10}-\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)

\(=\frac{1}{10}-\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{10}-\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{20}\right)\)

\(=\frac{1}{10}-\frac{1}{3}.\frac{3}{20}\)

\(=\frac{1}{10}-\frac{1}{20}=\frac{2}{20}-\frac{1}{20}=\frac{1}{20}\)

Đặt A= \(\frac{1}{10}-\frac{1}{40}-..-\frac{1}{340}\)

A=\(\frac{1}{10}-\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\right)\)

3A= \(\frac{1}{10}-(\frac{3}{5.8}+...+\frac{3}{17.20})\)

3A=\(\frac{1}{10}-\left(\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)

3A=\(\frac{1}{10}-\left(\frac{1}{5}-\frac{1}{20}\right)\)

3A=\(\frac{1}{10}-\frac{3}{20}\)

3A=\(-\frac{1}{20}\)

A=\(-\frac{1}{60}\)

ttiikk nha bạn

4 tháng 3 2020

\(=\frac{1}{2.5}-\frac{1}{5.8}-...-\frac{1}{17.20}.\)

\(=\frac{1}{10}-\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right).\)

\(=\frac{1}{10}-\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right).\)

\(=\frac{1}{10}-\frac{1}{3}\left(\frac{1}{5}-\frac{1}{20}\right)\)

\(=\frac{1}{10}-\frac{1}{3}.\frac{3}{20}\)

\(=\frac{1}{20}\)

14 tháng 1 2017

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}-3x=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)-3x=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)-3x=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)

\(\Rightarrow\frac{4949}{19800}-3x=\frac{451}{8120}\)

\(\Rightarrow3x=\frac{4949}{19800}-\frac{451}{8120}\)

\(\Rightarrow x=\left(\frac{4949}{19800}-\frac{451}{8120}\right):3\)

29 tháng 7 2018

\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)

\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)

\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3x+2}-\frac{1}{3x+5}\right)=\frac{4}{25}\)

\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3x+5}\right)=\frac{4}{25}\)

\(\frac{1}{2}-\frac{1}{3x+5}=\frac{12}{25}\)

\(\frac{1}{3x+5}=\frac{1}{50}\)

=> 3x+5 = 50

3x = 45

x = 15

7 tháng 9 2016

hề gặp bạn oy

 

7 tháng 9 2016

ò

bài kho quá à

19 tháng 9 2017

Ta có: 

\(-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

đặt \(A=1+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

   \(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}\)

\(A-\frac{1}{2}A=\frac{1}{2}A\Rightarrow A=\frac{1-\frac{1}{2^{11}}}{\frac{1}{2}}=2-\frac{1}{2^{10}}\)

19 tháng 9 2017

\(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

\(=-1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

Đặt  \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

\(A=1-\frac{1}{1024}=\frac{1023}{1024}\)

Vậy, \(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}=-1-A=-1-\frac{1023}{1024}=-\frac{2047}{1024}\)

16 tháng 8 2016

đề có sai k đó bạn

16 tháng 8 2016
Cho tam giac AbA
26 tháng 6 2016

a) A = 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 + 1/14.17 + 1/17.20

=> 3A = 1/2 - 1/5 + 1/5 - .... + 1/14 - 1/17 + 1/17 - 1/20

=> 3A = 1/2 - 1/20 = 9/20

=> A = 3/20

b) 200410 + 20049 = 20049(1+2004) = 20049 . 2005

200510  = 20059  . 2005

Do 20059 > 20049 nên 200410 + 20049 < 200510

19 tháng 1 2018

\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}\cdot\frac{998}{1000}\)

\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{499}{1000}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{499}{1000}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{499}{1000}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{499}{1000}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{499}{1000}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{1000}\)

=>x+1=1000

=>x=999